Question 16.10: Derive the frequency equation for uniform beam with elastic ......

Expressions for displacement and slope at x=l

\begin{aligned}& y(l)=Y_0 S(k l)+\frac{\theta_0}{k} T(k l)+\frac{Q_0}{k^3 E I} V(k l)=0, \\& \theta(l)=Y_0 k V(k l)+\theta_0 S(k l)+\frac{Q_0}{k^2 E I} U(k l)=0\end{aligned}            (a)

Initial displacement and shear force are connected as Y_0=-Q_0 / c ; the negative sign means the Y_0  and Q_0  are directed opposite. Therefore formulas (a) may be rewritten as

\begin{aligned}& \frac{T(k l)}{k} \theta_0+\left[\frac{V(k l)}{k^3 E I}-\frac{S(k l)}{c}\right] Q_0=0  , \\& S(k l) \theta_0+\left[\frac{U(k l)}{k^2 E I}-\frac{k V(k l)}{c}\right] Q_0=0\end{aligned}             (b)

We obtained two algebraic homogeneous equations with respect to unknown \theta_0  and Q_0 . The frequency equation is D=0, where D is determinant of the coefficients of unknown initial parameters \theta_0  and \theta_0 .

D=\left|\begin{array}{cc}\frac{T(k l)}{k} & \frac{V(k l)}{k^3 E I}-\frac{S(k l)}{c} \\S(k l) & \frac{U(k l)}{k^2 E I}-\frac{k V(k l)}{c}\end{array}\right|=0 .

In the expanded form the frequency equation becomes

\frac{S^2-T V}{T U-S V}=-\frac{c}{k^3 E I}

Multiplication of this equation by factor \lambda^3=k^3 l^3  leads to the frequency equation in terms of Krylov-Dunkan functions

\lambda^3 \frac{S^2(\lambda)-T(\lambda) V(\lambda)}{T(\lambda) U(\lambda)-S(\lambda) V(\lambda)}=-\frac{c l^3}{E I}

Frequency equation in the elementary functions is

\lambda^3 \frac{\cosh \lambda \cos \lambda+1}{\cosh \lambda \sin \lambda-\sinh \lambda \cos \lambda}=-\frac{c l^3}{E I}

The root \lambda  of this equation allows calculating the frequency of free vibration \omega=k^2 \sqrt{\frac{E I}{m}}=\frac{\lambda^2}{l^2} \sqrt{\frac{E I}{m}} .

Special Cases

1. Let c=0. In this case we have free-clamped beam. The frequency equation is

S^2(\lambda)-T(\lambda) V(\lambda)=0

In the elementary function \cosh \lambda \cos \lambda+1=0 . The roots of this equations are

\lambda_1=1.8751, \quad \lambda_2=4.6941, \cdots

The frequencies of free vibration are

\omega_1^2=\frac{\lambda_1^2}{l^2} \sqrt{\frac{E I}{m}}=\frac{1.8751^2}{l^2} \sqrt{\frac{E I}{m}}, \quad \ldots

2. Let c=\infty . In this case we have pinned-clamped beam. The frequency equation is

T(\lambda) U(\lambda)-S(\lambda) V(\lambda)=0

In the elementary function we have \cosh \lambda \sin \lambda-\sinh \lambda \cos \lambda=0 \quad  or \quad \tan \lambda=\tanh \lambda . The roots of this equation are \lambda_1=3.9266, \quad \lambda_2=7.0685, \cdots . The frequencies of free vibration are

\omega_1^2=\frac{\lambda_1^2}{l^2} \sqrt{\frac{E I}{m}}=\frac{3.9266^2}{l^2} \sqrt{\frac{E I}{m}}, \quad \ldots

Numerical solution of the frequency equation is presented in Table 16.5.